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3 years ago

13

What is the maximum volume of 0.25 M sodium hypochlorite solution (NaOCl, laundry bleach) that can be prepared by dilution of 1.

00 L of 0.80 M NaOCl?

2 answers:

ehidna [41]3 years ago

8 0

Answer:

0.313L

Explanation:

0.25Mx1.00L=0.80Mx V2

so, 0.25x1.00/0.80=0.313L

Iteru [2.4K]3 years ago

7 0

Answer: 3.2L

Explanation:

C1 = 0.80M

V1 = 1L

C2 = 0.25M

V2 =?

C1V1 = C2V2

0.8 x 1 = 0.25 x V2

V2 = (0.8 x 1) /0.25 = 3.2L

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Propose the structure of the following HNMR data.

Varvara68 [4.7K]

Answer:

1-Chloropropane is likely the answer (attached a picture)

Explanation:

First off there are 3 peaks and 3 carbons which indicates to me that this will be a chain without any symmetry and that each carbon has hydrogens on it.

Second the triplet at 1.0 that integrates to 3 likely correlates to a CH3 (methyl) group. Peaks are very upfield triplets that integrate to three are almost always methyl peaks.

Third the triplet at 3.7 is indicative of being next to the halogen. Hydrocarbons by themselves do not have peaks that far downfield meaning that its shift could only be explained by the chlorine being involved. Also we know that this can't be next to the methyl group since its multiplicity is to low to be next to it.

That leaves the multiplet at 1.75 being the hydrogens on the middle carbon which also makes sense since it is more downfield then the methyl group (due to being closer to the chlorine) but is not far enough downfield to say the chlorine is there. It also makes sense that it is a multiple since it would be a hextet due to the adjacent 5 hydrogens which can't always be resolved.

I hope this helps and let me know if anything is unclear or needs further explanation.

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Anna71 [15]

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The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. Combustion of

gavmur [86]

<u>Answer:</u> The empirical formula for the given compound is $C_3H_6O$

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Conversion factor: 1 g = 1000 mg

Mass of $CO_2=6.32mg=0.00632g$

Mass of $H_2O=2.58g=0.00258g$

Mass of compound = 2.78 mg = 0.00278 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.00632 g of carbon dioxide, $\frac{12}{44}\times 0.00632=0.00172g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 0.00258 g of water, $\frac{2}{18}\times 0.00258=0.000286g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.00278) - (0.00172 + 0.000286) = 0.000774 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.00172g}{12g/mole}=1.43\times 10^{-4}moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.000286g}{1g/mole}=2.86\times 10^{-4}moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.000774g}{16g/mole}=4.83\times 10^{-5}moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $4.83\times 10^{-5}mol$

For Carbon = $\frac{1.43\times 10^{-4}}{4.83\times 10^{-5}}=2.96\approx 3$

For Hydrogen = $\frac{2.86\times 10^{-4}}{4.83\times 10^{-5}}=5.92\approx 6$

For Oxygen = $\frac{4.83\times 10^{-5}}{4.83\times 10^{-5}}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 6 : 1

Hence, the empirical formula for the given compound is $C_3H_{6}O_1=C_3H_6O$

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Nady [450]

Answer:

1.0809

Explanation:

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